# Simplifying Exponents (exponent laws) [closed]

So I have this equation:

$${\left(\left(\frac{w}{x}\right)^{\frac{y}{x}}\right)}^{\frac{z}{x}}$$

And I know that

$${\left(a^b\right)}^c = a^{bc}$$

So I figured I'd simplify it into this:

$$\left(\frac{w}{x}\right)^{\frac{y}{x}\frac{z}{x}}$$

And then further into this:

$$\left(\frac{w}{x}\right)^{\frac{yz}{x^2}}$$

But for some reason, it doesn't appear to be working in all cases. What did I do wrong?

EDIT: In response to the Unclear what you're asking close votes, I am requesting assistance in finding the error in my simplification.

• It's not true that $a^{b^c}=a^{b\cdot c}$, for instance, take $a=3,b=2,c=3$, we have $a^{b^c}=2^8=256$ but $a^{b\cdot c}=2^6=64$.
– user265675
Dec 17, 2015 at 1:42
• Your use of $x^{y^z}$ is ambiguous. It may mean $x^{\left(y^z\right)}$ or $\left(x^y\right)^z$. Most mathematicians use the former sense but you seem to be using it in the latter sense Dec 17, 2015 at 1:44
• Dec 17, 2015 at 1:45
• @Henry that's probably my problem. I am evaluating it from left to right Dec 17, 2015 at 1:46
• Also $\dfrac{y}{x} \times \dfrac{z}{x} \not = \dfrac{yz}{2x}$ unless $x=2$ or $yz=0$ Dec 17, 2015 at 1:47

In general, a calculator will provide one answer to $a^{\frac1b}$, when there are actually $b$ existing roots.
• Then it depends if the calculator does $x^a$ or $x^\frac1a$ first. Doing roots second produces more answers. Doing roots second produces less answers. Turning the whole fraction into decimals should give only one answer. Dec 17, 2015 at 2:04